(a) (i) An AM signal s(t) = A_c[1 + kₐ m(t)]cos(2πf_c t) is applied to the system shown in the figure. Show that the message…

(a) (i) Use cos²x=(1+cos2x)/2. Then s(t)² = A_c²[1+k_a m(t)]² cos²(2πf_c t) = (A_c²/2)[1+k_a m(t)]² + (A_c²/2)[1+k_a m(t)]² cos(4πf_c t). The squarer output v_1(t) contains a DC term, a message term proportional to k_a m(t), a message-squared term proportional to k_a² m(t)², and the same terms at 2f_c. Because m(t) is limited to |f|≤ω, [1+k_a m(t)]² is limited to |f|≤2ω. The second term is centred at 2f_c. Since f_c>2ω, a low-pass filter with cutoff f_L satisfying 2ω<f_L<2f_c removes the 2f_c term and gives v_2(t)=(A_c²/2)[1+k_a m(t)]². Since |k_a m(t)|<1, the bracket is positive, so the square-rooter is exact and v_3(t)=√(v_2(t))=(A_c/√2)[1+k_a m(t)]. Thus m(t)=(√2 v_3(t)/A_c - 1)/k_a. This is a DC offset and scale. A_c is in V, k_a is dimensionless, f_c and ω are in Hz, v_2 is in V², and v_3 is in V.

(a) (ii) Use the quadrature-envelope method. Write s(t)=A_c cosω_c t - βA_c sinω_m t sinω_c t = X cosω_c t + Y sinω_c t, with X=A_c and Y=-βA_c sinω_m t. For f_c≫f_m, the envelope is E(t)=√(X²+Y²)=A_c√(1+β² sin²ω_m t). Thus E_max=A_c√(1+β²), E_min=A_c, and R(β)=E_max/E_min=√(1+β²). For 0≤β≤0.4, plot R versus β as y=√(1+β²); it is monotonic because dR/dβ=β/√(1+β²)≥0, and it passes through (0,1), (0.2,1.0198), and (0.4,1.0770). The ratio is not constant unless β=0; the curve is almost linear over this small range, but the exact expression should be used. For power, expand s(t)². The cross term averages to zero, and <sin²ω_m t>=<sin²ω_c t>=1/2, so P=(A_c²/2)(1+β²/2). The unmodulated carrier power is P_0=A_c²/2. Hence P/P_0=1+β²/2, percentage = (100+50β²)%; for β=0.4 it is 108%. The extra 50β²% is carried by the quadrature sideband term. If A_c is in V across 1 Ω, powers are in W.

(b) (i) Square-wave inverters have a fixed fundamental and large low-order odd harmonics. The fundamental rms is about 0.9 V_DC and cannot be reduced without reducing V_DC. The 3rd harmonic is one-third of the fundamental, which is a major problem for induction motors; they also have high THD, large filters, torque ripple, and poor power factor. PWM inverters control fundamental amplitude by dimensionless modulation index, push dominant harmonics to f_s, reduce THD, allow smaller filters, and improve current waveform and power factor. Spectrum:

• Square wave: lines at f_1, 3f_1, 5f_1, 7f_1 with relative rms 1, 1/3, 1/5, 1/7.
• Bipolar PWM: fundamental at f_1; clusters near multiples of f_s. Around f_s the sidebands are f_s±2f_1, f_s±4f_1, f_s±6f_1, ...; f_s±f_1 is absent.
• Unipolar PWM: fundamental at f_1; clusters near multiples of f_s. Around f_s the sidebands are f_s±f_1, f_s±2f_1, f_s±3f_1, ... . Unipolar switching uses +V_DC, 0, -V_DC, 0, so each voltage step is V_DC rather than 2V_DC, giving lower harmonic amplitudes than bipolar PWM. PWM amplitudes depend on modulation index and f_s/f_1, but the sideband locations above are the key difference.

(b) (ii) For the square wave, v(θ)=V_DC for 0<θ<π and -V_DC for π<θ<2π. The sine coefficient is b_n=(2/π)∫_0^π V_DC sin nθ dθ=(2V_DC/π n)(1-cos nπ). Thus b_n=0 for even n and b_n=4V_DC/(π n) for odd n. The alternating signs do not affect rms values. The rms value is V_n,rms=|b_n|/√2=(4V_DC)/(π n√2). The full-wave rms is V_DC, and the fundamental is 0.9 V_DC; the remaining power is in harmonics. With V_DC=400 V and f_1=50 Hz, taking the next two prominent harmonic orders as 3rd and 5th:

• Fundamental, 50 Hz: V_1,rms=1600/(π√2)=360.2 V.
• Third, 150 Hz: V_3,rms=120.1 V.
• Fifth, 250 Hz: V_5,rms=72.0 V. For the standard voltage-cancellation waveform, α is the angle by which voltage is cancelled at each end of a half-cycle; α=0 reduces to the square wave. The output is zero for 0<θ<α and π-α<θ<π, and +V_DC for α<θ<π-α; the negative half-cycle is symmetric. For odd n, b_n=(2/π)∫_α^(π-α) V_DC sin nθ dθ =(2V_DC/π n)(-cos n(π-α)+cos nα). For odd n, cos(nπ-nα)=-cos nα, so b_n=(4V_DC)/(π n)cos nα. Hence V_n,rms=(4V_DC)/(π n√2)|cos nα|. With α=20°:
• Fundamental: V_1,rms=360.2 cos20°=338.5 V.
• Third: V_3,rms=120.1 cos60°=60.0 V.
• Fifth: V_5,rms=72.0 |cos100°|=12.5 V. The 5th is small because cos100° is small, showing the cancellation effect. The 7th would be 39.4 V, but the next two harmonic orders are 3rd and 5th.

(c) (i) Use T(s)∝R_2 s/(R_2²+s²X_2²). Let s_m=R_2/X_2. Then T(s)∝s_m s/(s_m²+s²). Differentiating with respect to s and setting dT/ds=0 gives s=s_m, so T_max occurs at s_m. Also T_max∝1/2 and T_st∝s_m/(1+s_m²). Given T_st=0.75T_fl and T_max=2.5T_fl, T_max/T_st=10/3=(1+s_m²)/(2s_m). Thus 3s_m²-20s_m+3=0, so s_m=(10±√91)/3. The torque curve has two roots for a torque below T_max; the normal motoring maximum is the low-slip root below 1: s_m=(10-√91)/3=0.1535, i.e. 15.35%.

(c) (ii) Let full-load slip be s_f. Then T_max/T_fl=(s_m²+s_f²)/(2s_m s_f)=2.5. Thus s_f²-5s_m s_f+s_m²=0, so s_f=s_m(5±√21)/2. The larger root lies above s_m and is not the normal full-load slip. The full-load point is on the low-slip side, so s_f=s_m(5-√21)/2=0.0320, i.e. 3.20%. Check: T_fl/T_max=0.4 and T_st/T_fl=0.75, consistent with the data. The hp, voltage, and frequency data are not needed because the torque ratios fix the slips.